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C. MEX Game 1：
题目大意：
思路解析：
重要的是那种只有一个的数字，因为如果这个数字有两个及以上，那么我可以再鲍勃删除之后，再拿，也一定能拿得到，所以瓶颈是只有一个的数字，如果这样的数字有多个，那我们只能选择最小的那个。然后循环所有数字，看我们最小的拿不到的数字是那个，这个数字就是答案。
代码实现：
import java.io.*;
import java.math.BigInteger;
import java.util.*;public class Main {static int[] a = new int[200005];public static void main(String[] args) throws IOException {int t = f.nextInt();for (int o = 0; o < t; o++) {int n = f.nextInt();for (int i = 0; i <= n; i++) {a[i] = 0;}for (int i = 0; i < n; i++) {int num = f.nextInt();a[num]++;}int f = 0;for (int i = 0; i <= n; i++) {if (a[i] == 0) {w.println(i); break;}else {if (a[i] == 1){if (f == 1) {w.println(i); break;}else f = 1;}}}}w.flush();w.close();br.close();}static PrintWriter w = new PrintWriter(new OutputStreamWriter(System.out));static Input f = new Input(System.in);static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));static class Input {public BufferedReader reader;public StringTokenizer tokenizer;public Input(InputStream stream) {reader = new BufferedReader(new InputStreamReader(stream), 32768);tokenizer = null;}public String next() {while (tokenizer == null || !tokenizer.hasMoreTokens()) {try {tokenizer = new StringTokenizer(reader.readLine());} catch (IOException e) {throw new RuntimeException(e);}}return tokenizer.nextToken();}public String nextLine() {String str = null;try {str = reader.readLine();} catch (IOException e) {// TODO 自动生成的 catch 块e.printStackTrace();}return str;}public int nextInt() {return Integer.parseInt(next());}public long nextLong() {return Long.parseLong(next());}public Double nextDouble() {return Double.parseDouble(next());}public BigInteger nextBigInteger() {return new BigInteger(next());}}
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